Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 21 (Algebraic Number Theory)

An additive valuation on a field is a map $v:K^{\times }\to \Gamma$, where $\Gamma$ is a totally ordered abelian group, with $v(xy)=v(x)+v(y)$ and $v(x+y)\ge \min (v(x),v(y))$ whenever $x+y\ne 0$. Set $v(0)=\infty$. We consider nontrivial valuations; if the trivial valuation is allowed, it must be listed separately from the asserted classification. Two valuations are equivalent when their valuation rings agree, or equivalently their ordered value-group images identify in a way compatible with the maps. Real-valued equivalent nontrivial valuations differ by positive scaling.

For a valuation on $\mathbb{Q}$, every integer has nonnegative value, because it is a sum of copies of $1$ or its negative. If all prime numbers had value zero, unique factorization would make the valuation trivial. Thus some prime $p$ has $v(p)>0$. There cannot be two such primes: for distinct $p,q$, Bézout's identity gives $ap+bq=1$ with integers $a,b$, and the valuation inequality would give $v(1)>0$. All other prime values therefore vanish. Factoring the numerator and denominator of a rational number givesThe image is the cyclic ordered group generated by $v(p)>0$, proving equivalence to the P-adic valuation. This is the classification of nontrivial valuations on the rational numbers; an Archimedean absolute value is not an additive valuation satisfying this ultrametric inequality.

The simple-root form of Hensel's lemma says that for a complete discrete valuation ring $\mathcal{O}$ with maximal ideal $\mathfrak{p}$, a polynomial $f\in \mathcal{O}[X]$ and $a\in \mathcal{O}$ with $f(a)\equiv 0(\mathrm{mod}\mathfrak{p})$ and $f^{\prime }(a)\not\equiv 0(\mathrm{mod}\mathfrak{p})$ have a unique root in $a+\mathfrak{p}$.

For $X^3+1$ over ${\mathbb{Q}}_7$, all roots are integral: a negative valuation would make the leading term the unique term of lowest valuation. Modulo $7$, the roots are $3,5,6$, and $3X^2$ is nonzero at each. Hensel's lemma gives three distinct lifts, and a cubic has no further roots. The number is $3$.

For $X^2+2X+4$ over ${\mathbb{Q}}_2$, a root is again integral and reduces to zero modulo $2$. Write it as $2y$ with $y\in {\mathbb{Z}}_2$. The equation becomes $y^2+y+1=0$, impossible modulo $2$. The number is $0$.

For $3X^3+X+3$ over ${\mathbb{Q}}_3$, a putative root of valuation $k<0$ gives term valuations $1+3k,k,1$. The first is uniquely minimal, a contradiction. Thus all roots are integral. Reduction modulo $3$ is $X$, whose root zero is simple since the derivative $9X^2+1$ is a unit everywhere. Hensel's lemma gives exactly one lift. The number is $1$.

Normalize the discrete valuations so a uniformiser has value one, and write $k_K,k_L$ for the residue fields. The ramification index $e$ is specified by $v_L{|}_K=e{v}_K$, and the residue degree is $f=[k_L:k_K]$. For finite extensions of complete discretely valued fields, $[L:K]=ef$. One way to see the degree equality is that ${\mathcal{O}}_L$ is a finite free ${\mathcal{O}}_K$-module of rank $[L:K]$; reduction modulo a uniformiser of $K$ has $e$ successive quotients isomorphic to $k_L$, hence dimension $ef$ over $k_K$.

An unramified extension has $e=1$ and separable residue extension, equivalently $[L:K]=f$ with separable residue extension. A totally ramified extension has $f=1$, equivalently $[L:K]=e$. The separability condition in the unramified extension definition matters if the residue field is imperfect; it is automatic for finite residue fields.

Suppose $L/K$ is unramified. Choose a primitive element of a field extension $\bar{x}$ for the finite separable extension $k_L/k_K$ and lift it to $x\in {\mathcal{O}}_L$. Since the residue degree of $K(x)/K$ is at least $[k_K(\bar{x}):k_K]=[L:K]$, necessarily $K(x)=L$. Its monic minimal polynomial of an algebraic element $f_x$ has coefficients in ${\mathcal{O}}_K$. Its reduction has degree $[L:K]$ and annihilates $\bar{x}$, whose minimal polynomial over $k_K$ has that same degree. The two coincide, so ${\bar{f}}_x$ is separable.

Conversely, suppose $L=K(x)$, $x\in {\mathcal{O}}_L$, and ${\bar{f}}_x$ is separable. It must be irreducible: otherwise its coprime factors lift by the factorization form of Hensel's lemma, contradicting irreducibility of $f_x$. Hence $\bar{x}$ has degree $[L:K]$ over $k_K$. The degree equality forces $f=[L:K]$ and $e=1$, and the residue extension is separable. We have proved the unramified generator criterion with separable reduction:

Now let $k_L={\mathbb{F}}_q$, $q=p^f$. Every element of $k_L^{\times }$ is a simple root of $X^{q-1}-1$, whose derivative is a unit at each root. Hensel's lemma lifts each of these $q-1$ elements uniquely to a root in ${\mathcal{O}}_L^{\times }$. Thus $L$ contains all $q-1$ roots of $X^{q-1}-1$; these are the nonzero Teichmuller lifts.

More generally, a root of unity of order prime to $p$ reduces injectively into $k_L^{\times }$. Indeed, if such a root reduces to $1$, uniqueness in Hensel's lemma for $X^m-1$ makes it equal to $1$. Consequently every prime-to-$p$ order divides $q-1$.

A root whose order has a nontrivial $p$-part produces a primitive $p$th root of unity $\zeta$. It reduces to $1$ in characteristic $p$. Sinceall the factors have the same positive integral valuation. HenceThis is the ramification bound for a primitive pth root of unity. ThereforeFor $p=2$ the hypothesis $e<1$ cannot occur, so that instance is vacuous rather than a claim excluding the ever-present root $-1$.

For a finite Galois extension of local fields, let $G=\mathrm{Gal}(L/K)$, $G_{-1}=G$, and, for integers $i\ge 0$, define the lower ramification groups byIn particular, $G_0$ is the inertia group. The identity satisfies every bound because $v_L(0)=\infty$.

If the extension is totally ramified, its residue fields agree. Choose representatives in ${\mathcal{O}}_K$ of their common residue field. Every $a\in {\mathcal{O}}_L$ has a convergent uniformiser expansion $a={\sum }_{j\ge 0}{c}_j{\pi }_L^j$ with these representatives $c_j$, all fixed by $G$. For $j\ge 1$,Each summand in the second factor has valuation $j-1$. Thus a bound $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$ implies the same bound on $\sigma (a)-a$, by convergence and the valuation inequality. Necessity follows by testing $a={\pi }_L$. This proves the uniformizer criterion for lower ramification groups:

Put $\zeta ={\zeta }_{p^n}$, with $n\ge 1$. The shifted cyclotomic polynomial ${\Phi }_{p^n}(1+X)$ is Eisenstein at $p$: its constant term is $p$, and modulo $p$ it is $X^{p^{n-1}(p-1)}$. Therefore $L/{\mathbb{Q}}_p$ is totally ramified of degree $p^{n-1}(p-1)$ and $\pi =\zeta -1$ is a uniformiser.

The Galois group is $(\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }$, with ${\sigma }_a(\zeta )={\zeta }^a$. For $a\ne 1$, let $r=v_p(a-1)$, $0\le r<n$. Then ${\zeta }^{a-1}$ is a primitive $p^{n-r}$th root. Its difference from one is a uniformiser in the corresponding smaller cyclotomic field, and the relative ramification index is $p^r$. HenceThe uniformizer criterion for lower ramification groups now determines every group. Write $U_j=\{a\in (\mathbb{Z}/p^n\mathbb{Z}{)}^{\times }:a\equiv 1(\mathrm{mod}{p}^j)\}$, with $U_n=\{1\}$. Then the lower ramification filtration of a prime-power cyclotomic extension isFor $n=1$ the middle range is empty and the extension is tame. The formula also includes $p=2$; then $U_1$ is already the whole group, so the first possible drop is later than in the odd-prime case. For $p=2,n=1$, the entire extension is trivial.

Write $\alpha =\sqrt[3]{3}$ and $\zeta ={\zeta }_3$. The cubic is Eisenstein and adjoining $\zeta$ adds at most a quadratic extension, so $[L:{\mathbb{Q}}_3]\le 6$. SetUsing ${\zeta }^2+\zeta +1=0$ and ${\alpha }^3=3$, compute ${\pi }^3=1+2\zeta$ and ${\pi }^6=-3$. Thus $\pi$ satisfies the Eisenstein polynomial $X^6+3$. This proves $L={\mathbb{Q}}_3(\pi )$, degree six, with $\pi$ a uniformiser and the extension totally ramified. It is the splitting field of $X^3-3$, so its Galois group is $S_3$. This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

The six automorphisms have $\sigma (\alpha )={\zeta }^a\alpha$, $a\in \{0,1,2\}$, and $\sigma (\zeta )={\zeta }^b$, $b\in \{1,2\}$. Since $v_L(3)=6$, we have $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$.

If $b=1$ and $a\ne 0$, then $\sigma (\pi )={\zeta }^{-a}\pi$, whence $v_L(\sigma (\pi )-\pi )=1+3=4$. These are the two nonidentity elements of the cyclic subgroup $C_3$.

If $b=2$, use ${\zeta }^2-1=(\zeta -1)(\zeta +1)=-(\zeta -1){\zeta }^2$. Then $\sigma (\pi )=-{\zeta }^{2-a}\pi$. The coefficient $-{\zeta }^{2-a}-1$ reduces to $-2\ne 0$ in the residue field ${\mathbb{F}}_3$, so the difference has valuation one. These three automorphisms are the transpositions.

The uniformizer criterion for lower ramification groups consequently givesAs a consistency check, the different exponent is $(6-1)+3(3-1)=11$. The derivative of the monogenic Eisenstein polynomial gives the same value, $v_L(6{\pi }^5)=6+5=11$. In particular, stopping the wild filtration at $G_1$ would give the wrong different.

A place of a number field is an equivalence class of nontrivial absolute values on a field. Its finite places correspond to nonzero prime ideals of the ring of integers of a number field; its infinite places come from real embeddings and conjugate pairs of complex embeddings.

For a embedding of a number field into a p-adic algebraic closure $\iota :K\hookrightarrow {\overline{\mathbb{Q}}}_p$, pull back the p-adic absolute value. This gives a place above $p$. Every element of $\mathrm{Gal}({\overline{\mathbb{Q}}}_p/{\mathbb{Q}}_p)$ preserves that absolute value, by uniqueness of its extension to each finite local extension, so equivalent embeddings give the same place.

Conversely, a place $v$ above $p$ gives a completion of a number field at a prime ideal $K_v$, a finite extension of ${\mathbb{Q}}_p$. Embed it into ${\overline{\mathbb{Q}}}_p$ and restrict to $K$. If two embeddings give the same place, they extend to two ${\mathbb{Q}}_p$-embeddings of this same completion. They are conjugate under the absolute Galois group: their finite separable images lie in a finite normal closure, and an isomorphism of such images extends to an automorphism of the algebraic closure. The completion really is the closure of the embedded $K$, since rational coefficients are dense in the ${\mathbb{Q}}_p$-span of a primitive element. Thus we obtain the embedding orbit description of finite places:

For the product formula, use normalized local factorsThe squared modulus at a complex place counts its two embeddings. Equivalently one can use ordinary complex modulus and put exponent two in the product. These normalized local factors for the product formula must be specified; arbitrary representatives of place classes would not satisfy the unweighted printed formula.

The fractional principal ideal $(x)$ has only finitely many nonzero prime exponents. Its ideal norm is $|N_{K/\mathbb{Q}}(x)|$: for integral $x$, multiplication by $x$ on an integral basis has this determinant in absolute value, equal to the index of $(x)$ in ${\mathcal{O}}_K$; a quotient gives the fractional case. Hence the finite-place product is $|N_{K/\mathbb{Q}}(x){|}^{-1}$. The infinite-place product is $|N_{K/\mathbb{Q}}(x)|$, by the embedding expression for the field norm. Therefore all but finitely many local factors are one, and

Finally, let ${\Delta }_K\ne 0$ be the integer field discriminant, defined by the determinant of the trace pairing on an integral basis. If a rational prime $p$ ramifies, the algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ has a nonzero nilpotent element: use prime ideal factorization and the Chinese remainder theorem to see a nonzero nilpotent in a factor ${\mathcal{O}}_K/{\mathfrak{p}}^e$ with $e>1$. If $y$ is nilpotent, multiplication by $yz$ is nilpotent for every $z$ in this commutative algebra, and therefore has trace zero. Thus $y$ is in the radical of its trace pairing, making the reduced discriminant zero. We have proved the discriminant obstruction to ramification:Only finitely many rational primes divide this nonzero integer, so only finitely many primes ramify.

The idele group is the multiplicative restricted productIts restricted product topology has basic open sets ${\prod }_{v\in S}{U}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains every infinite place, and $U_v$ is open in $K_v^{\times }$. In particular, the restriction to almost all local unit groups is part of the topology, not merely the unrestricted product topology.

The diagonal map $K^{\times }\to J_K$ is injective and well-defined by finite support of the principal ideal's valuations; its image consists of principal ideles. To prove discreteness of principal ideles, take the neighbourhood of $1$ which requires $x_v\in {\mathcal{O}}_v^{\times }$ at every finite place and $|x_v-1|<1/2$ in the ordinary real or complex modulus at every infinite place. A diagonal element in it is a global unit, so $x-1\in {\mathcal{O}}_K$. If $x\ne 1$, the nonzero integer $N_{K/\mathbb{Q}}(x-1)$ has absolute value at least one, but the archimedean bounds make its absolute value less than $2^{-[K:\mathbb{Q}]}$. This contradiction proves $K^{\times }$ is a discrete subgroup of $J_K$.

For a modulus of a number field, write $\mathfrak{m}$ for its finite ideal part together with any selected real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals coprime to the finite part. Let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$ locally at each finite place in the modulus, and $a>0$ at each selected real place. The ray class group isEquivalently it is the quotient $J_K/(K^{\times }{U}_{\mathfrak{m}})$, with the corresponding principal-unit factors at finite places and positive multiplicative factors at selected real places. The question now assumes that no infinite places occur.

Put $R_{\mathfrak{m}}={\prod }_{v\in S}{\mathcal{O}}_v^{\times }/(1+{\pi }_v^{m_v}{\mathcal{O}}_v)$. Global units map to $R_{\mathfrak{m}}$ by reduction. To map $R_{\mathfrak{m}}$ to the ray class group, choose, by the Chinese remainder theorem, an integral $a$ having the prescribed unit residues at all $v\in S$, and send the tuple to the ray class of $(a)$. A different choice changes the ratio by an element congruent to one at every such place, so the map is well-defined and multiplicative.

The forgetful map to the ordinary ideal class group is surjective: weak approximation for number fields multiplies an ideal by a principal ideal so that all its valuations in $S$ vanish. Its kernel consists exactly of classes represented by principal ideals coprime to $S$, and their generators supply the local unit residues. Finally, a residue tuple maps to the identity ray class precisely when its chosen $a$ differs from a congruence-one generator by a global unit. Thus the ray class exact sequence isThe leftmost arrow need not be injective, explaining the absence of an initial zero.

For $K=\mathbb{Q}(\sqrt{2})$, the ring of integers of Q of square root two is $\mathbb{Z}[\sqrt{2}]$. Its absolute field norm is a Euclidean function. Indeed, for $a+b\sqrt{2}\in K$, choose nearest integers $m,n$; writing $u=a-m$, $w=b-n$, both bounded by $1/2$, gives $|u^2-2w^2|\le 1/2<1$. Applying this to a quotient proves Euclidean division, hence the ring is a principal ideal domain and $\mathrm{Cl}(K)=0$. This is norm-Euclideanity of the integers adjoined square root two.

Write $\mathfrak{p}=(\sqrt{2})$ and $\mathfrak{q}=(1+2\sqrt{2})$. Their ideal norms are $2$ and $7$, respectively; ${\mathfrak{p}}^2=(2)$ and the residue field at $\mathfrak{q}$ is ${\mathbb{F}}_7$, with $\sqrt{2}\equiv 3$ because $1+2\sqrt{2}\equiv 0$. The finite local quotient product for the requested modulus isThe first factor consists of $1$ and $1+t$.

The given fundamental unit $ϵ=1+\sqrt{2}$ and $-1$ generate the global units. Their images areThe first image has order six; its cube is $(1+t,1)$ and its square is $(1,2)$, generating the first factor and the order-three subgroup of the second. The second image supplies the missing order-two element of ${\mathbb{F}}_7^{\times }$. Thus the global unit map is surjective onto all $12$ local unit residues. By the exact sequence, or the unit-surjectivity criterion for a trivial finite ray class group,There is no positivity condition here: the modulus has no infinite part, so using the unit $-1$ is legitimate and essential in the surjectivity calculation.